Question

如何在python 2.4中逐月添加以下时间戳，即第3个月中发生的所有时间戳应该加在一起。

例如：结果应该像2011-03总时间戳是1:00:45，依此类推其他月份..

2011-03-07  0:27:41
2011-03-06  0:13:41
2011-03-08  0:17:40
2011-03-04  0:55:40
2011-05-16  0:55:40
2011-05-18  0:55:40
2011-07-16  0:55:40
2011-07-17  0:55:40

Answer 1

这个怎么样：

import datetime
import re
from collections import defaultdict
months = defaultdict(int)
# months = {}  # for Python 2.4
with open("test.txt") as timestamps:
    for line in timestamps:
        month = line[:7]
        time = re.search(r"(\d+):(\d+):(\d+)", line)
        if time:
             seconds = int(time.group(1))*3600 + \
                       int(time.group(2))*60 +   \
                       int(time.group(3))
             months[month] += seconds
             # if month in months:          # Python 2.4
             #     months[month] += seconds
             # else:
             #     months[month] = seconds
for month in sorted(months.keys()):
    print("Times for {}: {}".format(month, 
                                    datetime.timedelta(seconds=months[month])))

输出：

Times for 2011-03: 1:54:42
Times for 2011-05: 1:51:20
Times for 2011-07: 1:51:20

Answer 2

假设数据已经排序，您可以使用itertools.groupby：

import datetime as dt
import itertools as it

data='''\
2011-03-07  0:27:41
2011-03-06  0:13:41
2011-03-08  0:17:40
2011-03-04  0:55:40
2011-05-16  0:55:40
2011-05-18  0:55:40
2011-07-16  0:55:40
2011-07-17  0:55:40
'''.splitlines()

dates=[dt.datetime.strptime(line,'%Y-%m-%d %X') for line in data]
for key,group in it.groupby(dates,lambda d: (d.year,d.month)):
    seconds=sum(date.hour*3600+date.minute*60+date.second for date in group)
    print('{k[0]}-{k[1]:02d} {d}'.format(
        k=key,
        d=dt.timedelta(seconds=seconds)))

产量

2011-03 1:54:42
2011-05 1:51:20
2011-07 1:51:20

并且（当然）如果数据尚未排序，那么您可以在使用dates.sort()之前使用itertools.groupby对其进行排序。

Answer 3

这两个解决方案--AFAIK - 都应该适用于任何2.x版本的python（从而保证了一些向后兼容性）

仅依赖于正则表达式库的实现：

import re
data = '''
2011-03-07  0:27:41
2011-03-06  0:13:41
2011-03-05  0:17:40
2011-03-04  0:55:40
2011-05-16  0:55:40
2011-05-16  0:55:40
2011-07-16  0:55:40
2011-07-16  0:55:40
'''
def group(month):
    li = re.findall(r'2011-%s-\d\d  (\d+:\d+:\d+)' % str(month).zfill(2), data)
    seconds = 0
    for log in li:
        log = [int(n) for n in log.split(':')]
        seconds += log[0]*3600 + log[1]*60 + log[2]
    hours = seconds / 3600
    seconds -= 3600*hours
    minutes = seconds / 60
    seconds -= 60*minutes
    return "%02d:%02d:%02d" % (hours, minutes, seconds)

for month in range(1,13):
    print "In month %d you worked %s" % (month, group(month))

这里的实施没有依赖：

def group(month):
    to_find = '2011-%s-' % str(month).zfill(2)
    logs = []
    for line in data.split('\n'):
        point = line.find(to_find)
        if point != -1:
            logs.append(line[-8:])
    seconds = 0
    for log in logs:
        log = [int(n) for n in log.split(':')]
        seconds += log[0]*3600 + log[1]*60 + log[2]
    hours = seconds / 3600
    seconds -= 3600*hours
    minutes = seconds / 60
    seconds -= 60*minutes
    return "%02d:%02d:%02d" % (hours, minutes, seconds)

在python中添加时间戳

3 个答案: